A solution of `KMnO_4` is reduced to various products depending upon its pH. At pH lt 7 it is reduced to a colourless solution (A), at pH = 7 it forms a brown precipitate (B) and at pH gt 7 it gives a green solution ( C), (A),(B) and( C) are

Options

A. `{:("(A)","(B)","(C)"),(Mn^(2+),MnO_2,MnO_4^(2-)):}`
B. `{:("(A)","(B)","(C)"),(MnO_2,mn^(2+),MnO_4^(2-)):}`
C. `{:("(A)","(B)","(C)"),(Mn^(2+),MnO_4^(2-),MnO_2):}`
D. `{:("(A)","(B)","(C)"),(MnO_4^(2-),Mn^(2+),MnO_2):}`

Answer

A

Explanation

At pH lt 7, in acidic medium
`MnO_4^(-)+8H^(+)+5e^(-) to underset(("brown precipitate"))(MnO_2)+4OH^(-)`
At pH gt 7, in alkaline medium
`MnO_4^(-)+e^(-)to underset(("green"))(MnO_4^(2-))`